# Sequence $f_n$ bounded in $L^1$ such that $\int_0^x f_n\to \int_0^x f$ for all $x,$ but such that …

Find a sequence $f_n,$ bounded in $L^1([0,1]),$ and $f\in L^1,$ such that $$\lim_{n\to \infty}\int_0^x f_n=\int_0^x f$$ for all $x\in [0,1],$ and such that$\{f_n\}$ does not converge weakly to $f$ in $L^1$

Below is my attempt:

let $$f_n(x)=2n\chi_{[0,1/n]}(x)$$ then $f_n \in L^1([0,1])$, for all $x$ with $\int_{[0,1]} f_n =2$

define $f$ to be identically $0$, ie. $f\equiv 0$.

then $f_n \to f$ pointwise a.e

So since $f_n \geq 0, \forall n$ and $f_n$ is increasing, by monotone convergence theorem $$\lim_{n\to \infty}\int_0^x f_n=\int_0^x f=\int_0^x 0 = 0$$

Below is Royden's definition of weak convergence

Let $E$ be measurable set, and $1 \leq p< \infty$ and $q$ a conjugate of $p$, then $\{f_n\}$ is said to converge weakly to $f$ in $L^p(E)$ if and only if $$\lim_{n\to \infty}\int_E g.f_n=\int_E g.f , \forall g \in L^q(E)$$

As per this, I define $g=\chi_{[0,1]}(x)$, then clearly $g\in L^{\infty}([0,1])$, but $$\lim_{n\to \infty}\int_0^x g\cdot f_n=\int_0^xf_n =2$$ and $$\lim_{n\to \infty}\int_0^x g\cdot f=\int_0^x 0=0$$

hence $f_n$ does not converge weakly to $f$ in $L^1([0,1])$

Is this meaningful? if not can someone help me come up with one that works. thank you

• The integral of $f_n$ equals $2$, I believe. Also, $*$ is reserved for convolution. If you want to emphasize a product, use a cdot: $f\cdot g$. Otherwise it looks correct to me. – uniquesolution Jun 22 '17 at 15:39
• @uniquesolution You're right I just corrected the integral – Cnine Jun 22 '17 at 15:47
• But I just realized my sequence is not Bounded as required , $\{f_n\}$ grows without bounds, if I'm right. Any suggestions on how to make it bouunded – Cnine Jun 22 '17 at 15:51
• It is bounded in $L^1$. Where else do you want it to be bounded? – uniquesolution Jun 22 '17 at 15:56
• I mean is there a constant $M$ such that for all $n$, $|f_n|\leq M$? – Cnine Jun 22 '17 at 16:04

We can choose $a_ 2 > a_3 >\cdots \to 0$ such that $(1/n-a_n,1/n+a_n), n=2,3, \dots$ are pairwise disjoint subintervals of $[0,1].$ For each such $n,$ define

$$f_n = \frac{1}{a_n}\chi_{(1/n, 1/n + a_n)} - \frac{1}{a_n}\chi_{(1/n-a_n, 1/n)}.$$

Then $\|f_n\|_1 = 2$ for all $n,$ and $\int_0^x f_n \to 0$ for each $x\in [0,1].$

However $f_n$ does not converge weakly to $0.$ To see this, define

$$g(x) = \sum_{n=2}^{\infty}\chi_{(1/n, 1/n + a_n)} - \chi_{(1/n-a_n, 1/n)}.$$

Then $g\in L^\infty[0,1],$ and $\int_0^1 f_n g =2$ for all $n.$

• To courageous downvoter: Is there a reason you did this? Or is this just a drive-by shooting? – zhw. Jun 23 '17 at 1:50
• is $||f_n||_1=2$ or $0$, Also is there anyway to redefine those intervals to make them easy to understand? I could not figure out why they are disjoint. – Cnine Jun 26 '17 at 13:40